Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+px+2=0$ and $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ are the roots of the equation $2x^2+2qx+1=0$, then $(\alpha -\frac{1}{\alpha })(\beta -\frac{1}{\beta })(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })$ is equal to

• A. $\frac{9}{4}(9+p^2)$
• B. $\frac{9}{4}(9+q^2)$
• C. $\frac{9}{4}(9-p^2)$
• D. $\frac{9}{4}(9-q^2)$

Answer 1

The correct option is C $\frac{9}{4}(9-p^2)$

$\alpha$ and $\beta$ are the roots of the equation $x^2+px+2=0$.
$\Rightarrow \alpha +\beta =-p,\alpha \beta =2$
$\frac{1}{\alpha }$ and $\frac{1}{\beta }$ are the roots of the equation $2x^2+2qx+1=0$
$\Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }=-q,\frac{1}{\alpha \beta }=\frac{1}{2}$
$\Rightarrow \frac{\alpha +\beta }{\alpha \beta }=-q\Rightarrow \frac{-p}{2}=-q$
$\Rightarrow p=2q$
Now
$(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })=\alpha \beta +\frac{1}{\alpha \beta }+2$
$=2+\frac{1}{2}+2=\frac{9}{2}$
and
$(\alpha -\frac{1}{\alpha })(\beta -\frac{1}{\beta })=\alpha \beta +\frac{1}{\alpha \beta }-\frac{\alpha }{\beta }-\frac{\beta }{\alpha }$
$=2+\frac{1}{2}-\left[\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }\right]$
$=\frac{5}{2}-\left[\frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }\right]$
$=\frac{5}{2}-\left[\frac{p^2-4}{2}\right]$
$=\frac{9-p^2}{2}$

$\therefore (\alpha -\frac{1}{\alpha })(\beta -\frac{1}{\beta })(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })=\left(\frac{9-p^2}{2}\right)\left(\frac{9}{2}\right)$
$=\frac{9}{4}(9-p^2)$